9781118230725.pdf

232 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Calculations:Linear momentum is also conserved along

thexaxis because there is no net external force acting on

the coconut and pieces along that axis. Thus we have

PixPfx, (9-49)

wherePix0 because the coconut is initially at rest. To

getPfx, we find the xcomponents of the final momenta,

using the fact that piece Amust have a mass of 0.50M

(M0.20M0.30M):

pfA,x0.50MvfA,

pfB,x0.20MvfB,x0.20MvfBcos 50,

pfC,x0.30MvfC,x0.30MvfCcos 80.

Equation 9-49 for the conservation of momentum along the

xaxis can now be written as

PixPfxpfA,xpfB,xpfC,x.

Then, with vfC5.0 m/s and vfB9.64 m/s, we have

0 0.50MvfA0.20M(9.64 m/s) cos 50

0.30M(5.0 m/s) cos 80,

from which we find

vfA3.0 m/s. (Answer)

Sample Problem 9.06 Two-dimensional explosion, momentum, coconut

Two-dimensional explosion:A firecracker placed inside a
coconut of mass M, initially at rest on a frictionless floor,
blows the coconut into three pieces that slide across the floor.
An overhead view is shown in Fig. 9-13a. Piece C, with mass
0.30M, has final speed vfC5.0 m/s.

(a) What is the speed of piece B, with mass 0.20M?

KEY IDEA

First we need to see whether linear momentum is con-
served. We note that (1) the coconut and its pieces form a
closed system, (2) the explosion forces are internal to that
system, and (3) no net external force acts on the system.
Therefore, the linear momentum of the system is conserved.
(We need to be careful here: Although the momentum of
the system does not change, the momenta of the pieces cer-
tainly do.)

Calculations:To get started, we superimpose an xycoordinate
system as shown in Fig. 9-13b, with the negative direction of the
xaxis coinciding with the direction of v:fA.The xaxis is at 80

Additional examples, video, and practice available at WileyPLUS

Figure 9-13Three pieces of an

exploded coconut move off in

three directions along a

frictionless floor. (a) An over-

head view of the event. (b) The

same with a two-dimensional

axis system imposed.

with the direction of and 50with the direction of.
Linear momentum is conserved separately along each
axis. Let’s use the yaxis and write

PiyPfy, (9-48)

where subscript irefers to the initial value (before the ex-
plosion), and subscript yrefers to the ycomponent of
or.
The component Piyof the initial linear momentum is
zero, because the coconut is initially at rest. To get an ex-
pression for Pfy, we find the ycomponent of the final linear
momentum of each piece, using the y-component version of
Eq. 9-22 (pymvy):

pfA,y0,

pfB,y0.20MvfB,y0.20MvfBsin 50,

pfC,y0.30MvfC,y0.30MvfCsin 80.

(Note that pfA,y0 because of our nice choice of axes.)
Equation 9-48 can now be written as

PiyPfypfA,ypfB,ypfC,y.

Then, with vfC5.0 m/s, we have

0  0 0.20MvfBsin 50(0.30M)(5.0 m/s) sin 80,

from which we find

vfB9.64 m/s9.6 m/s. (Answer)

(b) What is the speed of piece A?

Pf

:

Pi

:

v:fC v:fB

A

B

C

vfB

vfC

vfA

100 °

130 °

(a)

B

C

vfB

vfC

vfA

80 °

(b)

x

y

50 °

A

The explosive separation

can change the momentum

of the parts but not the

momentum of the system.